If your thermopile is not touching the Al tube surface, it is _deducing_ the Al's temperature from the radiation it receives ... by converting the radiated power that it absorbs into electric voltage (at some efficiency). The (electrical) Power to Heat the tube will eventually raise the tube's Temperature until its radiated power cancels the Heating Power [radiated Power proportional to T^4 for a black-body] ... unfortunately, almost all of the radiated Power that it loses will be emitted from the tubes' outside surface [shiny, not even close to a black-body]. Obtaining an infrared spectrum from a small confined space like the inside of a tube will be tricky ... with visible light you could use a fiber-optic to extract the light out to a grating - but typical fiber-optic materials absorb IR bands.

What were you hoping to extract?
You have two measurements of the same temperature, one direct and the other based on power radiated and an assumption of black body behaviour. Any discrepancy will tell you the extent to which it is not a perfect black body. To get more data, like power as a function of wavelength, you would need a more versatile sensor.

peak wavelength - which I believe I can get from Wiens displacement Law

That law assumes black body radiation. If it is black body, yes you can predict it from either of your temperature measurements, and they must be the same. But you did not need to do experiment to get that.
Surely the only experimental result you will get is the discrepancy - possibly zero - between the two measurements. If there is a discrepancy then neither temperature gives a reliable figure for the peak wavelength.

If your thermopile is not touching the Al tube surface, it is _deducing_ the Al's temperature from the radiation it receives ... by converting the radiated power that it absorbs into electric voltage (at some efficiency). The (electrical) Power to Heat the tube will eventually raise the tube's Temperature until its radiated power cancels the Heating Power [radiated Power proportional to T^4 for a black-body] ... unfortunately, almost all of the radiated Power that it loses will be emitted from the tubes' outside surface [shiny, not even close to a black-body]. Obtaining an infrared spectrum from a small confined space like the inside of a tube will be tricky ... with visible light you could use a fiber-optic to extract the light out to a grating - but typical fiber-optic materials absorb IR bands.

That law assumes black body radiation. If it is black body, yes you can predict it from either of your temperature measurements, and they must be the same. But you did not need to do experiment to get that.
Surely the only experimental result you will get is the discrepancy - possibly zero - between the two measurements. If there is a discrepancy then neither temperature gives a reliable figure for the peak wavelength.

As I understand it, there are devices that measure the power in some narrow wavelength range. Satellites use them. I have no idea whether such could be used in your set-up.

Yes I will only have a difference between the two temperature readings but I was hoping this could provide more information. The thermopile detector has a filter on it currently so for example if the temperature of the blackbody is at 90 degrees and the secondary temperature sensor reads 90 but the thermopile reads 80 - is there no way of extracting what type of wavelengths may have been blocked out by the filter?

This filter will be removed at some point in the future. The data sheet has this graph and I am confused how they worked out the transmission % at each wavelength because I would like to reproduce this graph and technique so I can test performance outside of the wavelengths shown on this graph. (image attached).

Ok, that helps. But I fail to see how a filter will succeed in maintaining a temperature difference in a confined space. The filter itself will get hot, reradiate, etc. The thermopile will be subjected to a distorted spectrum, but must still converge on the ambient temperature.

Ok, that helps. But I fail to see how a filter will succeed in maintaining a temperature difference in a confined space. The filter itself will get hot, reradiate, etc. The thermopile will be subjected to a distorted spectrum, but must still converge on the ambient temperature.

Presumably it also has some explanation of what is being plotted. It is impossible to interpret out of context.

Unfortunately it does not.

I have my thermopile infrared detector aimed at a black body. This blackbody is emitting at all wavelengths, however my thermopile detector has a filter - for example a filter to only allow 3-8 micro meter infrared radiation. So the plank law spectrum graph (irradiance against wavelength) has essentially been chopped to just allow radiation in the 3-8 micrometer range through the lens.

How does my temperature measurement (calculated from voltage signal output on the detector) relate to the intensity of radiation in the wavelength band 3-8?

Will I be seeing different temperatures with different filters? Will the thermopile detector not even obtain the true temperature of the blackbody because not all the radiation is getting through when using a filter - example if i set the blackbody to 100 degrees will I not read 100 the detector due to the filter because some radiation intensity has been removed by the filter. The filter removes radiation outside of the filter range so there is less radiation to heat the detector to produce the voltage which gives tempeature.

Yes, but I am not sure what to expect from a thermopile outside the body. As I understand it, a thermopile registers the difference in temperature between two points. Even without a filter, why should a thermopile sensor outside the body be at the same temperature as the body? Its temperature will be whatever makes it lose heat by radiation (or conduction) at the same rate that it gets heat from the black body.

Maybe you do not expect it to be at the same temperature, merely at a temperature that relates to that of the blackbody. Certainly filtering out wavelengths will reduce the power reaching the thermopile and hence its temperature.

Please confirm the physical set-up, i.e. where the thermopile sensor is in relation to the black body.

Your original description had it inside the tubular black body, so I could not understand how a filter would achieve much.

Yes, but I am not sure what to expect from a thermopile outside the body. As I understand it, a thermopile registers the difference in temperature between two points. Even without a filter, why should a thermopile sensor outside the body be at the same temperature as the body? Its temperature will be whatever makes it lose heat by radiation (or conduction) at the same rate that it gets heat from the black body.

Maybe you do not expect it to be at the same temperature, merely at a temperature that relates to that of the blackbody. Certainly filtering out wavelengths will reduce the power reaching the thermopile and hence its temperature.

Please confirm the physical set-up, i.e. where the thermopile sensor is in relation to the black body.

Sorry, The thermopile (a number of thermocouples connected in series) is pointing at the blackbody heater and they are 50cm distance apart.

To save anyone reading this from looking it up - A thermocouple has two 'layers', one 'hot' end exposed to where the radiation heat enters the sensor and the other 'cold' end placed on a heat sink to reduce its temperature (between the two layers is a low thermal conductivity layer to hinder heat transfer), a temperature difference between the two ends (layers) generates a voltage proportional to the temperature difference between them.

Next to the thermopile is a seperate simple temperature sensor. Currently the thermopile and temperature sensor give the same temperature readout in ambient room temperature - both approx 22 degrees without the blackbody. I can get both temperature and voltages for both sensors.

This again puzzles me. What do you mean by saying it is "pointing at" the black body? A thermopile is like a thermometer, it tells you its own temperature, not the temperature of something it is "pointed at". It does not really have a direction. If I point my face at a roaring fire, I feel my face getting warm, but it will be nowhere near the temperature of the fire. The temperature my face reaches is the point where the incoming radiation from the fire balances the rate at which heat is transported away from the face. The same will be true for the thermopile.

This is in contrast to the sensors used in satellites. They measure the incoming radiation power in a narrow wavelength range.

You originally wrote that the simple temperature sensor (thermometer?) is directly on the blackbody. Now you say it is next to the thermopile. If it is not directly on the black body then the same issue arises - it will tell you its own temperature, not that of the black body.

Since the two sensors will have different heat loss characteristics, when placed at a distance from the black body you cannot expect them to reach the same temperature. So there seems no point in having the thermometer at all.

Anyway, accepting all that, suppose you just use the thermopile and place different filters in the way. These will change the incoming power, so change the temperature reached. The problem is, how to deduce the power from the temperature? One way would be to calibrate the thermopile by raising the black body to various different temperatures with no filters. That would allow you to plot black body power output against thermopile reading. Then running it with filters you may be able to deduce how much power the filters are intercepting.